Obtaining The Mean From Two Sets Of Data

Prove that if set A, of size { n }_{ 1 }, has mean \overline { { x }_{ 1 } } and set B, of size { n }_{ 2 }, has a mean \overline { { x }_{ 2} } then the mean of the combined set of A and B is \overline { x } =\frac { { n }_{ 1 }\overline { { x }_{ 1 } } +{ n }_{ 2 }\overline { { x }_{ 2 } } }{ { n }_{ 1 }+{ n }_{ 2 } } .

——————

Set A:

{ q }_{ 1 },{ q }_{ 2 },{ q }_{ 3 },{ q }_{ 4 }

{ n }_{ 1 }=4

Therefore:

\overline { { x }_{ 1 } } =\frac { { q }_{ 1 }+{ q }_{ 2 }+{ q }_{ 3 }+{ q }_{ 4 } }{ 4 } =\frac { { q }_{ 1 }+{ q }_{ 2 }+{ q }_{ 3 }+{ q }_{ 4 } }{ { n }_{ 1 } }

Therefore:

{ n }_{ 1 }\overline { { x }_{ 1 } } ={ q }_{ 1 }+{ q }_{ 2 }+{ q }_{ 3 }+{ q }_{ 4 }

——————

Set B:

{ q }_{ 5 },{ q }_{ 6 },{ q }_{ 7 },{ q }_{ 8 },{ q }_{ 9 }

{ n }_{ 2 }=5

Therefore:

\overline { { x }_{ 2 } } =\frac { { q }_{ 5 }+{ q }_{ 6 }+{ q }_{ 7 }+{ q }_{ 8 }+{ q }_{ 9 } }{ 5 } =\frac { { q }_{ 5 }+{ q }_{ 6 }+{ q }_{ 7 }+{ q }_{ 8 }+{ q }_{ 9 } }{ { n }_{ 2 } }

Therefore:

{ n }_{ 2 }\overline { { x }_{ 2 } } ={ q }_{ 5 }+{ q }_{ 6 }+{ q }_{ 7 }+{ q }_{ 8 }+{ q }_{ 9 }

So:

\frac { \left( { q }_{ 1 }+{ q }_{ 2 }+{ q }_{ 3 }+{ q }_{ 4 } \right) +\left( { q }_{ 5 }+{ q }_{ 6 }+{ q }_{ 7 }+{ q }_{ 8 }+{ q }_{ 9 } \right) }{ 9 } \\ \\ =\frac { { n }_{ 1 }\overline { { x }_{ 1 } } +{ n }_{ 2 }\overline { { x }_{ 2 } } }{ 4+5 } \\ \\ =\frac { { n }_{ 1 }\overline { { x }_{ 1 } } +{ n }_{ 2 }\overline { { x }_{ 2 } } }{ { n }_{ 1 }+{ n }_{ 2 } }

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